Suppose $f$ is an non constant entire function with $ f(z) neq 0 $ for all $ z in C $. Then the set $ U = { z in C : | f(z) | <1 } $ has unbounded connected component.
My attempt
I have concluded some facts:
$ U $ is an open set;
$ f(z) neq 0 $ for all $ z in C $, then there exists $ epsilon >0 $ such that $ | f(z) | > epsilon$;
Since $f$ is non-vanishing, we can define another entire function g such that $ g(z) = frac{1}{ f(z) } $.
But I was not able to conclude the result.
Mathematics Asked by Learner_Shas on January 1, 2021
1 Answersnote that assuming $f$ non-constant ($f=2$ has $U$ empty), $g=1/f$ is non-constant and entire hence $g$ is unbounded, so $U$ being given by $|g(z)|>1$ must be unbounded (hence it has at least one unbounded connected component) as $V$ given by $|g(z)| le 1$ cannot contain the exterior of a disc radius $R$ for all $R$ large enough as that would imply $g$ bounded (as $g$ is bounded inside the closed disc of radius $R$, if it were bounded by $1$ outside, it would be bounded on the plane), while the union of $U$ and $V$ is the plane! Done!
Answered by Conrad on January 1, 2021
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